A THEOREM ABOUT LIMITS WITH APPLICATION TO MACLAURIN’S POLYNOMIAL OF CERTAIN
FUNCTIONS
LUCA GOLDONI
Abstract. In this article we consider a theorem for the limit of certain functions at x = 0. We shall prove that in some cases, the existence of the first derivative at x = 0 and the existence of a “functional equation” let us able to infer the existence of some derivatives of higher order.
1. The theorem
Theorem 1. Let be f : R → R a continuous function such that (a): There exist k > 1, a ∈ R, h ∈ Z+
such that f(kx) = kf (x) + a [f (x)]h ∀x ∈ R. (b): f (0) = 0. (c): There exists lim x→0 f(x) x = b. (d): The function g(x) = bx − f(x) xh
is bounded in a deleted neighborhood of x= 0. then ∃lim x→0 g(x) = L. where L= −a b k h 1 − 1 kh−1 −1 . Date: January 2, 2013.
2000 Mathematics Subject Classification. 26A06, 26A09.
Key words and phrases. Limits, elementary functions, elementary proof, Maclau-rin’s polynomials.
Dipartimento di Matematica. Universit`a di Trento.
Proof. Since g(x) is bounded in a deleted neighborhood of x = 0, we have that
−∞< λ= lim
x→0inf g(x) 6 limx→0sup g(x) = Λ < +∞.
Now, by hypothesis (a), we can write f(x) = kfx k + ahfx k ih for each x ∈ R, so that
g(x) = bx − f(x) xh = k x k − f x k kh x k h − a kh f x k h x k h Hence g(x) = g1(x) + g2(x). where g1(x) = 1 kh−1 x k − f x k x k h . and g2(x) = − a kh f x k h x k h . By hypotesis (c) it is lim
x→0inf g2(x) = limx→0sup g2(x) = limx→0g2(x) = −a
b k
h = A. Now, we remember that
lim
x→0inf g1(x) + limx→0inf g2(x) 6 limx→0inf (g1(x) + g2(x)) 6
6 lim
x→0sup (g1(x) + g2(x)) 6 limx→0sup g1(x) + limx→0sup g2(x).
(1) and so
lim
x→0inf g1(x) + limx→0inf g2(x) 6 λ 6 Λ 6 limx→0sup g1(x) + limx→0sup g2(x).
But lim x→0inf g1(x) = 1 kh−1λ (2) lim x→0sup g1(x) = 1 kh−1Λ. (3) so that 1 kh−1λ+ A 6 λ 6 Λ 6 1 kh−1Λ + A. It follows that A 1 − 1 kh−1 −1 6λ 6Λ 6 A 1 − 1 kh−1 −1 .
and so
λ= Λ = L = lim
x→0g(x).
This theorem can be used, at least in principle, in order to obtain the Maclaurin polynomials of certain functions without the use of the derivative of order higher than one. Indeed since
g(x) = bx − f(x)
xh = L + o(1) (x → 0)
we have
f(x) = bx − Lxh+ o(xh).
We get an example in order to clarify the previous statement. 2. An example
Le us consider f (x) = sin x. We remember that for any x ∈ R it is well known that
sin 3x = 3 sin x − 4 sin3
x.
Thus, we have that k = 1, a = −4, h = 3 and b = 1 because lim
x→0
sin x x = 1. Since, in this case
g(x) = x −sin x x3 .
is an even function, we can consider x > 0 only. In this case, since x <tan x in a deleted right neighborhood of x = 0, it follows that
g(x) = x −sin x x3 < tan x − sin x x3 . But lim x→0 tan x − sin x x3 = lim x→0 sin x x 1 − cos x x2 = 1 2.
thus g(x) is bounded in a right neighborhood of x = 0 and in a complete deleted neighborhood also. Hence, all hypotheses are satisfied and we have that lim x→0 x −sin x x3 = 4 1 3 3 1 − 1 32 −1 = 1 6. This means that
(4) sin x = x − 1
6x
3
+ o(x3
).
This is the Maclaurin’s polynomial of third order for the function sin x and we obtained it without use the derivatives of order higher than one.
Note 1. Perhaps the most important key point is that it is possible to write (5) g(x) = ckg x k + Fh(x) + sk,h(x) .
where x → sk,h(x) is a function of x such that lim
x→0sk,h(x) = Ak,h ∈ R. and Fh(x) = " f xk x k #h .
This is strongly related with the existence of a “functional equation” like
f(kx) = kf (x) + a [f (x)]h.
which can be said to be of order h. In principle, we can obtain the Maclaurin’s polynomial of a given order n if we are able to find the suitable functional equations for f . We get an example just to under-stand.
3. The previous example go further
For first we are going to see that there is a relation like (5). Since sin 5x = 5 sin x − 20 sin3
x+ 16 sin5 x. if we consider g(x) = x − 1 6x 3− sin x x5 . we can write g(x) = x − 1 6x 3 − 5 sinx 5 −20 sin 3x 5 x5 − 16 sin5 x 5 x5 .
Now, from (4), we have sin3 x= x −1 6x 3 + o(x3 ) 3 thus sin3 x= x3− 1 2x 5 + o(x5 ). Hence g(x) = x − 1 6x 3 − 5 sinx 5 −20 x 53 3− 1 2 x 55 5 x5 + o(1) − 16 sin5 x 5 x5 .
After some calculations, we have g(x) = h x 5 − 1 6 x 5 3− sin x 5 i 54 x 5 5 − 16 sin5 x 5 55 x 5 5 −10 1 55 + o(1). Thus, we have the relation (5) with
(x) = 1 54g x 5 −16 55 [F (x)] 5 + sk,h(x)
where sk,h(x) = −10 1 55 + o(1).
For second, we are going to prove that g(x) is bounded in a deleted neighborhood of x = 0. As before, it is enough to consider only a deleted right neighborhood of x = 0. If we consider
k(x) = sin x − x −1 6x 3 . we have that k(0) = 0 and
k′(x) = 1 − 1 2x 2− cos x thus k′(x) = sin 2 x cos x + 1− 1 2x 2
As before, from (4), we get sin2 x= x2−1 3x 4 + o x4 . so that k′(x) = x 2 2 1 − cos x (cos x + 1) −1 3 x4 cos x + 1+ o x 4 It follows that k′(x) = x 2 2 "x2 2 + o (x 2 ) (cos x + 1) # − 1 3 x4 cos x + 1 + o x 4 which leads to k′(x) = − 1 12 x4 cos x + 1+ o x 4 6 −x 4 24 + o x 4 < −x 4 48 Hence k(x) = x Z 0 k′(t)dt < − 1 48 x Z 0 t4 dt <0
in a suitable deleted right neighborhood of x = 0. In the same way we have that x −1 6x 3 −sin x > −x5.
This means that, even in this case, we have −∞ < λ 6 Λ < +∞ where, of course λ and Λ have the same meaning as before but they are related with the ne function g(x). In this case we get
1 625λ − 26 55 6λ 6Λ 6 1 625Λ − 26 55 so that lim x→0g(x) = − 1 120
and thus sin x = x − 1 6x 3 + 1 120x 5 + o(x5 ).
and this result has been obtained with no derivatives of order higher than one.
References
[1] G.H. Hardy “A course in Pure Mathematics” Cambridge University Press, 2004. [2] C.J. Tranter “Techniques of Mathematical analysis” Hodder and
Stoughton,1976.
Universit`a di Trento, Dipartimento di Matematica, v. Sommarive 14, 56100 Trento, Italy
